The present invention relates to a power supply unit for use by on-vehicle equipment.
Today, more and more people riding stereo-equipped vehicles are enjoying music on full blast using onboard subwoofers. To drive the subwoofers requires installing a large-output power amplifier.
This type of power amplifier consumes currents ranging from 20 at minimum to hundreds of amperes. For example, to drive 4.OMEGA. subwoofers on two channels (100 W, non-clip) requires furnishing a common linear amplifier (AB class) dissipating about 30 A.
FIG. 2 is a block diagram of a related art power amplifier meeting the above requirements. As illustrated, a power amplifier 40 placed inside of a trunk room 30 is powered by a battery 20 in an engine room 10 through wires "a" and "a'". In operation, the power amplifier 40 boosts the power from the battery 20 using a booster type DC--DC converter 42 and the boosted power drives a circuit load 44. Lines "b" and "b'" connect the booster type DC--DC converter 42 with the circuit load 44. The ground wire "a'" of the booster type DC--DC converter 42 is connected to vehicle body ground.
In the setup above, suppose that the battery voltage is 12V and that the wires "a" and "a'" carry a current of 30 A. In that case, if the battery voltage is boosted by the booster type DC--DC converter 42 by six times to 72V (or .+-.36V) and the conversion efficiency of the booster type DC--DC converter 42 is 100 percent, the current flowing through the wire "b" and "b'" is defined in the following equation: battery voltage.times.current flowing in wire "a"=boosted voltage.times.current flowing in wire "b". That is, there exists the relationship in which Ib denotes the current flowing in the wire "b": EQU 12(V).times.30(A)=72(V).times.Ib(A)
which means; EQU Ib(A)=5(A)
It can be seen that the current diminishes in inverse proportion to boosting ratios.
The related art setup above has some disadvantages. Specifically, an appreciably long distance between the battery 20 and the booster type DC--DC converter 42 necessitates a large current to flow in the extended wire "a". This involves a significant loss of current over the wire. Illustratively, if the wire "a" is a 10-meter-long vinyl-sheathed wire (equivalent to AWG 8) with a cross-sectional sectional area of 8 mm.sup.2, then a voltage drop of 0.7V occurs over the wire "a" which weighs as much as one kilogram or thereabout.
The larger the current, the greater the voltage drop entailing further loss of power. Where the length of the wire is fixed, there is only one way to forestall the voltage drop: increase the cross-sectional area of the wire. Inevitably, the wire becomes heavier than ever.